[tex]\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)-\mathbb P(A\cap B)[/tex]
Because [tex]A[/tex] and [tex]B[/tex] are independent, you have [tex]\mathbb P(A\cap B)=\mathbb P(A)\times\mathbb P(B)[/tex], so
[tex]\mathbb P(A\cup B)=0.60+0.39-0.60\times0.39=0.756[/tex]
If [tex]A[/tex] and [tex]B[/tex] are mutually exclusive, you have
[tex]\mathbb P(A\cup B)=\mathbb P(A)+\mathbb P(B)=0.60+0.39=0.99[/tex]
